When I first started entering blog giveaways, I noticed some bloggers posting something to the effect of, â€œI verify all entries.â€ Initially, I thought those bloggers must mean that once the random number generator picked a number, the blogger would verify that entry (and the corresponding mandatory entry if applicable) and disqualify it if it werenâ€™t valid. It turns out I was wrong. There are some bloggers out there that verify every, single entry before drawing a winner. This post is for them.
If youâ€™re verifying every entry before drawing a winner because you want to clean up spam and make your blog tidy, go for it. However, if you are verifying every entry because you think not doing so will change anyoneâ€™s chance of winning, donâ€™t bother. As long as your reaction to drawing an invalid entry is to redraw for a new winner (as opposed to just moving to the next person on the list), you arenâ€™t affecting anyoneâ€™s odds by leaving the invalid entries alone until you draw.
Consider the example where there are 100 entries (91 valid and 9 invalid), and Jane has 3 of the valid entries. I believe the confusion is because some people think that if the invalid entries are not deleted prior to drawing, Jane has only a 3 in 100 (3%) chance of winning, whereas if the invalid entries are all removed before the drawing, Jane has a 3 in 91 (3.3%) chance. The catch is that the 3 in 100 calculation does not take into account Janeâ€™s possibility of winning in a redraw if an invalid entry is drawn.
Letâ€™s work through a simple example. We have five entries: A (valid), B (valid), C (valid), D (invalid), and E (invalid). Weâ€™ll show that the odds of entry B winning the contest are 1/3 whether you leave the invalid entries in the pool or not. For the first drawing, all five entries have an equal chance of being drawn. Weâ€™ll visualize this as a bar split into five pieces like so:
If entry D (invalid) is chosen in the first drawing, we drop it from the entry pool and redraw. The entry pool for that second drawing is A, B, C and E, and all four entries have a 1/4 chance of being drawn. We visualize this by taking the grey bar for entry D in the first drawing and, without changing its size, split it into four equal pieces, representing each entry in the second drawing. Drawing entry E in the first round will result in a similar redraw:
Repeat a similar treatment for the third round:
If we plug all the redraws back into our original diagram, we get the distribution of every possible outcome of the contest:
Shuffling those pieces around to group them, we see that when we figure in the redraw results, each of the valid entries (A, B, and C) ends up with an equal chance (1/3) of winning the contest. We donâ€™t care how many draws were made, just whether the entry won in the end:
This example was simple enough to work out graphically, but the general case can be mathematically proven using convergent series to arrive at the same conclusion. Whether you remove the invalid entries ahead of time or not, the chance of any given valid entry winning the contest is one divided by the number of valid entries. The chance of a particular contestant winning the contest is the number of valid entries she submitted divided by the total number of valid entries in the contest.
In the five-entry example, for simplicity, the invalid entries dropped out one at a time as they were drawn. In a real drawing, you might drop several entries before you redraw. For example, if the entry you draw is invalid because the mandatory entry wasnâ€™t completed, you can delete all the rest of the entries by that contestant at the same time then do your redraw. The odds donâ€™t change.
The above information assumes that if you draw an invalid entry, you do a redraw. You can not simply move to the next entry on the list. Doing so opens your contest up to cheaters who put in invalid entries in order to increase the odds for the next person (e.g. a friend or second identity).